2. Найдите корни уравнения: 1) a) \(\frac{3x-9}{x-1} + \frac{x+6}{x+1} = 3\); б) \(\frac{4y+7}{2y-3} - \frac{y-3}{2y+3} = 1\); в) \(\frac{3}{x+2} - \frac{3}{2-x} = \frac{2}{x^2-4}\); г) \(\frac{2y-8}{y-5} + \frac{10}{y^2-25} = \frac{y+4}{y+5}\);

Решение: a) \(\frac{3x-9}{x-1} + \frac{x+6}{x+1} = 3\) ОДЗ: \(x
eq 1, x
eq -1\) \((3x-9)(x+1) + (x+6)(x-1) = 3(x-1)(x+1)\) \(3x^2 + 3x - 9x - 9 + x^2 - x + 6x - 6 = 3(x^2 - 1)\) \(4x^2 - x - 15 = 3x^2 - 3\) \(x^2 - x - 12 = 0\) \(D = (-1)^2 - 4(1)(-12) = 1 + 48 = 49\) \(x_1 = \frac{1 + \sqrt{49}}{2} = \frac{1+7}{2} = 4\) \(x_2 = \frac{1 - \sqrt{49}}{2} = \frac{1-7}{2} = -3\) Ответ: \(x_1 = 4, x_2 = -3\) б) \(\frac{4y+7}{2y-3} - \frac{y-3}{2y+3} = 1\) ОДЗ: \(y
eq \frac{3}{2}, y
eq -\frac{3}{2}\) \((4y+7)(2y+3) - (y-3)(2y-3) = (2y-3)(2y+3)\) \(8y^2 + 12y + 14y + 21 - (2y^2 - 3y - 6y + 9) = 4y^2 - 9\) \(8y^2 + 26y + 21 - 2y^2 + 9y - 9 = 4y^2 - 9\) \(6y^2 + 35y + 12 = 4y^2 - 9\) \(2y^2 + 35y + 21 = 0\) \(2y^2 + 35y + 21 = 0\) \(D = 35^2 - 4*2*30 = 1225 - 240 = 985\) \(y = \frac{-35 \pm \sqrt{985}}{4}\) Ответ: \(y_1 = \frac{-35 + \sqrt{985}}{4}, y_2 = \frac{-35 - \sqrt{985}}{4}\) в) \(\frac{3}{x+2} - \frac{3}{2-x} = \frac{2}{x^2-4}\) ОДЗ: \(x
eq \pm 2\) \(\frac{3}{x+2} + \frac{3}{x-2} = \frac{2}{(x-2)(x+2)}\) \(3(x-2) + 3(x+2) = 2\) \(3x - 6 + 3x + 6 = 2\) \(6x = 2\) \(x = \frac{1}{3}\) Ответ: \(x = \frac{1}{3}\) г) \(\frac{2y-8}{y-5} + \frac{10}{y^2-25} = \frac{y+4}{y+5}\) ОДЗ: \(y
eq \pm 5\) \(\frac{2y-8}{y-5} + \frac{10}{(y-5)(y+5)} = \frac{y+4}{y+5}\) \((2y-8)(y+5) + 10 = (y+4)(y-5)\) \(2y^2 + 10y - 8y - 40 + 10 = y^2 - 5y + 4y - 20\) \(2y^2 + 2y - 30 = y^2 - y - 20\) \(y^2 + 3y - 10 = 0\) \(D = 3^2 - 4(1)(-10) = 9 + 40 = 49\) \(y_1 = \frac{-3 + \sqrt{49}}{2} = \frac{-3+7}{2} = 2\) \(y_2 = \frac{-3 - \sqrt{49}}{2} = \frac{-3-7}{2} = -5\) Но \(y
eq -5\), следовательно, только \(y = 2\) Ответ: \(y=2\)